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The image formed by a concave mirror is found to be virtual, erect and enlarged. Where should the object be placed?
A concave mirror has a reflective surface that is curved inward and away from the light source. Concave mirrors reflect light inward to one focal point. Unlike convex mirrors, the image formed by a concave mirror shows different image types depending on the distance between the object and the mirror.
A ray of light falls on a plane mirror making an angle of 30° with the mirror. On reflection, the ray deviates through an angle of
∠AON = i = 30° = r = ∠ NOB
Ray AA´ deviates such that angle of
Deviation ∠ A´OB = 180° – 30° – 30° = 120°
The linear magnification produced by concave mirror is always positive. This is because
m = v/u. For a convex mirror, v is –ive, thus m becomes positive.
When two or more than two rays starting from a point on the object, after refraction through a lens, do not actually meet but appear to diverge from point, the image formed is
A concave mirror produces an image of 20 cm height which is five times magnified. The height of the object is
Given that u =  20 cm and m =  5
 20 =  v/5
v =  4
Thus, the height of the object is 4 cm.
The linear magnification of a convex mirror of focal length 15 cm is 1/3. The distance of the object from the mirror is
u = – 2 f
= – 2 × 15
= – 30 cm
The power of a convex lens P_{1} is equal to 4 D, which is placed in close contact with a concave lens having power P_{2} equal to – 10 D. What will be the power of the combination of the two lenses?
P = P_{1} + P_{2} = 4 D + (– 10 D) = – 6 D
The image of a small electric bulb fixed on the wall of a room is to be obtained, on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
u + v = 3 m (given)
As maximum distance between an object and image in case of a convex lens = 4 f
4 f = 3 m ⇒ f = 3/4 m = 0.75 m
The linear magnification of a convex lens is – 1 when object in front of the lens is
In a convex lens, when object is in front of the lens at 2 F_{1}, an inverted image of size of object (h_{2} = h_{1}) is formed at 2F_{2}. Therefore, m = – 1.
If a glass is placed in a liquid of refractive index that is equal to glass, it will
The refractive index of liquid must be greater than or equal to 1.5 in order to make the lens disappear.
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